Originally Posted by
thekrown I have points p, q, r,
p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)
Okay so I need to find the equation to the plane using these three points.
Step 1. I find vector pq and vector qr
vector pq = (2,1,2)
vector qr = (1,-2,-4)
Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)
Step 3. I sub this into any point using (I forget what it's called form) and I get
0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0
I think that's right, what do you think?